perplexus.info :: Just Math : A! + B! + C! = 3<sup>D</sup>

A! + B! + C! = 3^D (Posted on 2009-08-27)

Determine all possible quadruplet(s) (A, B, C, D) of nonnegative integer(s), with A < B < C, that satisfy this equation:

A! + B! + C! = 3^D

Submitted by K Sengupta

Rating: 5.0000 (1 votes)

Solution: (Hide)

(0,2,3,2), (0,2,4,3), (1,2,3,2) and (1,2,4,3) are the only possible solutions.

For an explanation, refer to the solution submitted by Daniel in this location.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
re: Metaphysics?	Charlie	2009-08-29 00:41:21
Metaphysics?	ed bottemiller	2009-08-28 21:02:36
Solution	Dej Mar	2009-08-28 11:18:06
Spreadsheet & Computer without proof	brianjn	2009-08-27 22:46:37
analytical solution	Daniel	2009-08-27 11:33:12

Please log in:

Login:
Password:
Remember me:

Sign up! \| Forgot password

Search:

Search body:

Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:


perplexus dot info